
An object kept on the principle axis is moving in the same direction as that of a mirror as shown in the figure. Speed of the object and mirror is $10m{s^{ - 1}}$ and $\dfrac{{40}}{{13}}m{s^{ - 1}}$ . Radius of the curvature of the mirror is $20cm$ . If the distance of the object from the mirror at this instant is $5xcm$ , the velocity of the image at this instant is found to be zero. Find x.

Answer
232.8k+ views
Hint: In this question the values of the speed of the object, the mirror, and the radius of curvature are already given. We need to use the relation between the speed of the image and the object to find the required solution.
Formula used:
${\vec V_{im}} = - \dfrac{{{v^2}}}{{{u^2}}}{\vec v_{om}}$
Complete answer:
Before starting the solution of the question let us write all the given values which are present in the question,
The velocity of the object = $10m{s^{ - 1}}$
The velocity of the mirror = $\dfrac{{40}}{{13}}m{s^{ - 1}}$
The radius of curvature of the mirror = $20cm$
To find = the distance of the object from the mirror at an instant of $5xcm$ and also to find the value of x.
As we start the solution by using the below formula,
${\vec V_{im}} = - \dfrac{{{v^2}}}{{{u^2}}}{\vec v_{om}}$
By putting all the values in the above formula, we get the result as,
$0 - \dfrac{{40}}{{13}} = - {\left( {\dfrac{{ - 10}}{{ - 10 + x}}} \right)^2}\left( {10 - \dfrac{{40}}{{13}}} \right)$
By doing further solution of the above equation we get,
$\dfrac{{ - 10}}{{ - 10 + x}} = \pm \dfrac{2}{3}$
By using the cross multiplication method, we get the following values,
$5x = 25, - 5$
As we know that the object is in front of the mirror,
Hence, $x = 5$ .
Therefore, the correct answer of $x$ is $5$ .
Note: Choose the formula wisely for solving this problem. Some does the mistake of using the mirror formula. Also, it is mentioned that If the distance of the object from the mirror at that instant is 5xcm, the velocity of the image at this instant is found to be zero. Thus, we just can’t put the value of the speed of the object and the mirror directly in the formula.
Formula used:
${\vec V_{im}} = - \dfrac{{{v^2}}}{{{u^2}}}{\vec v_{om}}$
Complete answer:
Before starting the solution of the question let us write all the given values which are present in the question,
The velocity of the object = $10m{s^{ - 1}}$
The velocity of the mirror = $\dfrac{{40}}{{13}}m{s^{ - 1}}$
The radius of curvature of the mirror = $20cm$
To find = the distance of the object from the mirror at an instant of $5xcm$ and also to find the value of x.
As we start the solution by using the below formula,
${\vec V_{im}} = - \dfrac{{{v^2}}}{{{u^2}}}{\vec v_{om}}$
By putting all the values in the above formula, we get the result as,
$0 - \dfrac{{40}}{{13}} = - {\left( {\dfrac{{ - 10}}{{ - 10 + x}}} \right)^2}\left( {10 - \dfrac{{40}}{{13}}} \right)$
By doing further solution of the above equation we get,
$\dfrac{{ - 10}}{{ - 10 + x}} = \pm \dfrac{2}{3}$
By using the cross multiplication method, we get the following values,
$5x = 25, - 5$
As we know that the object is in front of the mirror,
Hence, $x = 5$ .
Therefore, the correct answer of $x$ is $5$ .
Note: Choose the formula wisely for solving this problem. Some does the mistake of using the mirror formula. Also, it is mentioned that If the distance of the object from the mirror at that instant is 5xcm, the velocity of the image at this instant is found to be zero. Thus, we just can’t put the value of the speed of the object and the mirror directly in the formula.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding the Electric Field of a Uniformly Charged Ring

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

